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State true or false. Explain your reasoning. (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may be a 3-digit number. (iv) The cube of a 2-digit number may have seven or more digits. (v) Cube numbers have an odd number of factors.
(i) False – Odd³ = Odd (e.g., 3³ = 27) (ii) False – 2³ = 8 ends in 8 (iii) True – 4³ = 64 (2-digit → 2-digit), 5³ = 125 (2-digit → 3-digit) (iv) True – 99³ = 970299 (2-digit cube has 6 digits) (v) False – Only cube numbers that are also squares have odd number of factors. Others have even. FoRead more
(i) False – Odd³ = Odd (e.g., 3³ = 27)
(ii) False – 2³ = 8 ends in 8
(iii) True – 4³ = 64 (2-digit → 2-digit), 5³ = 125 (2-digit → 3-digit)
(iv) True – 99³ = 970299 (2-digit cube has 6 digits)
(v) False – Only cube numbers that are also squares have odd number of factors. Others have even.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessWhat number will you multiply by 1323 to make it a cube number?
1323 = 3³ × 7². For it to be a perfect cube, each prime must occur in multiples of 3. We already have 7², so we need one more 7 to make it 7³. Therefore, multiply 1323 by 7 to get 3³ × 7³ = (3×7)³ = 21³ = 9261 Answer: Multiply by 7 and cube root of the result is 21. For more NCERT Solutions fRead more
1323 = 3³ × 7². For it to be a perfect cube, each prime must occur in multiples of 3.
We already have 7², so we need one more 7 to make it 7³.
Therefore, multiply 1323 by 7 to get 3³ × 7³ = (3×7)³ = 21³ = 9261
Answer: Multiply by 7 and cube root of the result is 21.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessFind the cube roots of 27000 and 10648.
To find cube roots: 27000 = 3 × 3 × 3 × 10 × 10 × 10 = (3×10)³ = 30³ ⇒ √³27000 = 30 10648 = 2 × 2 × 2 × 11 × 11 × 11 = (2×11)³ = 22³ ⇒ √³10648 = 22 So, cube roots are 30 and 22 respectively. Both numbers are perfect cubes. For more NCERT Solutions for Class 8 Mathematics Ganita Prakash ChapteRead more
To find cube roots:
27000 = 3 × 3 × 3 × 10 × 10 × 10 = (3×10)³ = 30³ ⇒ √³27000 = 30
10648 = 2 × 2 × 2 × 11 × 11 × 11 = (2×11)³ = 22³ ⇒ √³10648 = 22
So, cube roots are 30 and 22 respectively. Both numbers are perfect cubes.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessCompute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?
When we calculate differences of perfect cubes, we do: 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125 First level: 7, 19, 37, 61 Second level: 12, 18, 24 Third level: 6, 6 The third-level differences are constant. This reveals that perfect cubes follow a third-degree pattern. So, the third successive diRead more
When we calculate differences of perfect cubes, we do:
1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125
First level: 7, 19, 37, 61
Second level: 12, 18, 24
Third level: 6, 6
The third-level differences are constant. This reveals that perfect cubes follow a third-degree pattern. So, the third successive differences of cubes are always equal.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessIs 500 a perfect cube?
Let’s factorise 500: 500 = 2² × 5³ For a number to be a cube, all prime factors must appear in triplets. Here, the factor 2 appears only twice, which cannot make a group of three. Therefore, 500 is not a perfect cube, as it cannot be written as the cube of any integer. For more NCERT SolutionRead more
Let’s factorise 500:
500 = 2² × 5³
For a number to be a cube, all prime factors must appear in triplets. Here, the factor 2 appears only twice, which cannot make a group of three. Therefore, 500 is not a perfect cube, as it cannot be written as the cube of any integer.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See less